Quote:
Originally Posted by carpetpool
Now we see that the discriminant D of S(x, n) = 2^r where r = (n+1)*2^n1

Prove it.
Quote:
Originally Posted by carpetpool
Since the exponent r is odd, each prime factor q dividing S(x, n) has the form k*2^(n+1)+1.

This is very false, it fails even for n=3: 2 divides S(3), but you can't write 2 in the k*2^4+1 form.